So i have a matrix A:
\begin{pmatrix} i & i\bar i \\ 1-i &\bar i \end{pmatrix}
How can i calculate the eigenvalues of a complex matrix like this? I already know how to do if the matrix is not complex, but in complex field?
Using the standard method for R field i got:
$p(A)$$=x^2-2ix+i$
So $P(x)$=$0$ for:
$x=i+2i^2=i-4$ ...and...
$x=i-2i^2=i+4$
Wich seems pretty clear results. Is this method legit? And if i want to calculate the relative eigenspaces of this eigenvalues i just need to use the same method of the R field?
$\endgroup$ 12 Answers
$\begingroup$Yes. All of the calculations are equivalent in all fields, not just $\mathbb{C}$ or $\mathbb{R}$.
$\endgroup$ 3 $\begingroup$Note $\;i \bar i=1$, so t the characteristic polynomial is $$\begin{vmatrix}i-\lambda&1\\1-i&-i-\lambda\end{vmatrix}=\lambda^2+i,$$ so the eigenvalues are the square roots of $-i$.
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